NLM BYU 24

Physics Solution Question 24

Solution to Question 24

Step 1: Analysis of the Wedge’s Motion

Since the wedge is made of a “very light material,” we can consider its mass to be negligible ($m_{wedge} \approx 0$). Consequently, the net horizontal force acting on the wedge must be zero at all times; otherwise, it would have infinite acceleration.

Let the wedge accelerate to the right with acceleration $a$. We will analyze the motion relative to the accelerating wedge (non-inertial frame). In this frame, we must apply a pseudo-force equal to $mass \times a$ directed to the left on both blocks.

Step 2: Force Analysis on the Wedge

The wedge experiences normal forces from both blocks. Let $N_1$ be the normal force from block $M$ and $N_2$ be the normal force from block $m$.

The horizontal component of the force exerted by block $M$ on the wedge is $N_1 \sin\theta$ (directed to the right). The horizontal component of the force exerted by block $m$ on the wedge is $N_2 \sin\theta$ (directed to the left).

Since the wedge is massless ($m_{wedge} \to 0$), the net horizontal force must be zero:

$$N_1 \sin\theta – N_2 \sin\theta = m_{wedge} a = 0$$ $$\Rightarrow N_1 = N_2$$

Step 3: Calculating Normal Forces

We analyze the forces perpendicular to the inclined planes for both blocks in the frame of the wedge.

For block $M$ (right side), the forces perpendicular to the incline are the component of gravity $Mg \cos\theta$ and the component of the pseudo-force $Ma \sin\theta$. Since there is no motion perpendicular to the plane:

$$N_1 + Ma \sin\theta = Mg \cos\theta$$ $$N_1 = M(g \cos\theta – a \sin\theta)$$

For block $m$ (left side), the pseudo-force $ma$ acts to the left. Its component perpendicular to the plane pushes the block into the wedge, adding to the normal force:

$$N_2 = mg \cos\theta + ma \sin\theta$$

Step 4: Finding the Acceleration ($a$)

Equating the expressions for $N_1$ and $N_2$ derived above:

$$m(g \cos\theta + a \sin\theta) = M(g \cos\theta – a \sin\theta)$$

Rearranging to solve for $a$:

$$mg \cos\theta + ma \sin\theta = Mg \cos\theta – Ma \sin\theta$$ $$ma \sin\theta + Ma \sin\theta = Mg \cos\theta – mg \cos\theta$$ $$a \sin\theta (M + m) = g \cos\theta (M – m)$$ $$a = g \cot\theta \left( \frac{M – m}{M + m} \right)$$

Step 5: Condition for Block $m$ to Slide Up

For the block $m$ to start sliding up the incline relative to the wedge, the net force along the incline in the upward direction must be positive.

The forces acting along the incline are:

  • Component of pseudo-force pushing up the incline: $ma \cos\theta$
  • Component of gravity pulling down the incline: $mg \sin\theta$

The condition for upward sliding is:

$$ma \cos\theta > mg \sin\theta$$ $$a \cos\theta > g \sin\theta$$ $$a > g \tan\theta$$

Step 6: Final Calculation

Substitute the expression for acceleration $a$ into the inequality:

$$g \cot\theta \left( \frac{M – m}{M + m} \right) > g \tan\theta$$ $$\frac{M – m}{M + m} > \frac{\tan\theta}{\cot\theta}$$ $$\frac{M – m}{M + m} > \tan^2\theta$$

Cross-multiply (since $M+m > 0$):

$$M – m > (M + m) \tan^2\theta$$ $$M – m > M \tan^2\theta + m \tan^2\theta$$ $$M – M \tan^2\theta > m + m \tan^2\theta$$ $$M (1 – \tan^2\theta) > m (1 + \tan^2\theta)$$

Isolating $m$:

$$m < M \frac{1 - \tan^2\theta}{1 + \tan^2\theta}$$

Using the trigonometric identity $\cos 2\theta = \frac{1 – \tan^2\theta}{1 + \tan^2\theta}$, we get the final range:

$$m < M \cos 2\theta$$
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